Gain insights into algorithms and their implementation for problem-solving in Python. Learn about recursion, dynamic programming, greedy algorithms, and graph algorithms to enhance coding proficiency and problem-solving skills.

As a developer, mastering the concepts of algorithms and being proficient in implementing them is essential to improving problem-solving skills. This course aims to equip you with an in-depth understanding of algorithms and how they can be utilized for problem-solving in Python.

Starting with the basics, you'll gain a foundational understanding of what algorithms are, with topics ranging from simple multiplication algorithms to analyzing algorithms. Then, you’ll delve into more advanced topics like recursion, backtracking, dynamic programming, and greedy algorithms. You'll also learn about basic graph algorithms, depth-first search, shortest paths, minimum spanning trees, and all-pairs shortest paths. 

By the end of this course, you'll have acquired a wide range of skills that will significantly enhance your ability to solve problems efficiently in Python. Through this course, not only will you improve your coding skills, but you will also gain confidence in tackling complex problems.

Mastering Algorithms for Problem Solving in Python

# Introduction to storing files

Suppose we have a set of $n$ files that we want to store on a magnetic tape. In the future, users will want to read those files from the tape. Reading a file from tape isn’t like reading a file from a disk; first, we have to fast-forward past all the other files, and that takes a significant amount of time. Let $L[1 .. n]$ be an array listing the lengths of each file; specifically, file $i$ has length $L[i]$. If the files are stored in order from $1$ to $n$, then the cost of accessing the $k$th file is:

$$
cost(k) = \underset{i=1}{\overset{k}\sum}L[i]
$$

# Expected cost of accessing a random file
The cost reflects the fact that before we read file $k$, we must first scan past all the earlier files on the tape. If we assume for the moment that each file is equally likely to be accessed, then the expected cost of searching for a random file is:

$$
E[cost] = \underset{k=1}{\overset{n}{\sum}}\frac{cost(k)}{n} = \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} \space \underset{i=1}{\overset{k}{\sum}}L[i]
$$

If we change the order of the files on the tape, we change the cost of accessing the files; some files become more expensive to read, but others become cheaper. Different file orders are likely to result in different expected costs. Specifically, let $π(i)$ denote the index of the file stored at position $i$ on the tape. Then the expected cost of the permutation $π$ is:

$$
E[cost(\pi)] = \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} \space \underset{i=1}{\overset{k}{\sum}}L[\pi(i)]
$$

# Finding the best order to store files on tape
Which order should we use if we want this expected cost to be as small as possible? The answer seems intuitively clear: sort the files by increasing length. But intuition is a tricky business.

**Lemma 1.** $E[cost(π)]$ is minimized when $L[π(i)] ≤ L[π(i + 1)]$ for all $i$.

**Proof:** Suppose $L[π(i)] > L[π(i + 1)]$ for some index $i$. To simplify notation, let $a=π(i)$ and $b=π(i+1)$. If we swap files $a$ and $b$, then the cost of accessing $a$ increases by $L[b]$, and the cost of accessing $b$ decreases by $L[a]$. Overall, the swap changes the expected cost by $(L[b]− L[a])/n$. But this change is an improvement because $L[b] < L[a]$. Thus, if the files are out of order, we can decrease the expected cost by swapping some unordered pairs of files.


This is our first example of a correct greedy algorithm. To minimize the total expected cost of accessing the files, we put the file that is cheapest to access first and then recursively write everything else; no backtracking, no dynamic programming, just make the best local choice and blindly plow ahead. If we use an efficient sorting algorithm, the running time is clearly $O(n \log n)$ plus the time required to actually write the files. To show that the greedy algorithm is actually correct, we've proved that the output of any other algorithm could be improved by some sort of exchange.

Let’s generalize this idea further. Suppose we are also given an array $F [1 .. n]$ of access frequencies for each file; file $i$ will be accessed exactly $F[i]$ times over the lifetime of the tape. Now the total cost of accessing all the files on the tape is:

$$
\sum cost(\pi) = \underset{k=1}{\overset{n}{\sum}} \begin{pmatrix}
 &  F[\pi(k)] . \underset{i=1}{\overset{k}{\sum}} L[\pi(i)]\\
\end{pmatrix} = \underset{k=1}{\overset{n}{\sum}} \space \underset{i=1}{\overset{k}{\sum}}L[\pi(i)]
$$

As before, reordering the files can change this total cost. So what order should we use if we want the total cost to be as small as possible? (This question is similar in spirit to the optimal binary search tree problem, but the target data structure and the cost function are both different, so the algorithm must be different, too.)


We already proved that if all the frequencies are equal, we should sort the files by increasing size. If the frequencies are all different, but the file lengths $L[i]$ are all equal, then intuitively, we should sort the files by decreasing access frequency with the most-accessed file first. In fact, this is not hard to prove (hint, hint) by modifying the proof of Lemma 1. But what if the sizes and the frequencies both vary? In this case, we should sort the files by the ratio $L/F$.

**Lemma 2.** $\sum cost(\pi)$ is minimized when $\frac{L[\pi(i)]}{F[\pi(i)]} \leq \frac{L[\pi(i+1)]}{F[\pi(i+1)]}$ for all $i$

**Proof:** Suppose $L[π(i)]/F[π(i)] > L[π(i + 1)]/F[π(i + i)]$ for some index $i$. To simplify notation, let $a = π(i)$ and $b = π(i+1)$. If we swap files $a$ and $b$, then the cost of accessing $a$ increases by $L[b]$, and the cost of accessing $b$ decreases by $L[a]$. Overall, the swap changes the total cost by $L[b]F[a] − L[a]F[b]$. But this change is an improvement because

$$
\frac{L[a]}{F[a]} >  \frac{L[b]}{F[b]} \Leftrightarrow L[b]F[a] - L[a]F[b] <0.
$$

Thus, if any two adjacent files are out of order, we can improve the total cost by swapping them.

Understand the different techniques used to solve the storing files on tape problem efficiently.

Getting Started

Introduction to Algorithm

Recursion

Backtracking

Dynamic Programming

Greedy Algorithms

Prove Your Skills: A Five-Chapter Assessment

Basic Graph Algorithms

Depth-First Search

Minimum Spanning Trees

Shortest Paths

All-Pairs Shortest Paths

Pushing Your Limits: A Comprehensive Assessment

Wrapping up

Storing Files on Tape

Introduction to storing files